A $5 \times 8$ rectangle can be rolled to form two different cylinders with different maximum volumes. What is the ratio of the larger volume to the smaller volume? Express your answer as a common fraction.
Solution: Holding the rectangle vertically, we can form a cylinder with height 8 and base circumference of 5.  Let this cylinder have volume $V_A$ and radius $r_A$; we have $2\pi r_A = 5$ so $r_A = \frac{5}{2\pi}$ and $V_A = \pi r_A ^2 h = \pi \left(\frac{5}{2\pi}\right)^2 (8) = \frac{50}{\pi}$.

Holding the rectangle horizontally, we can form a cylinder with height 5 and base circumference of 8.  Similarly, let this cylinder have volume $V_B$ and radius $r_B$; we have $2\pi r_B = 8$ so $r_B = \frac{4}{\pi}$ and $V_B = \pi r_B^2 h = \pi \left(\frac{4}{\pi}\right)^2 (5) = \frac{80}{\pi}$.

Hence, the ratio of the larger volume to the smaller volume is $\frac{80/\pi}{50/\pi}=\boxed{\frac{8}{5}}$.